# Prove that $\frac{1}{2}\cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n+1}}$ without induction

Prove that $\frac{1}{2}\cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n+1}}$

I know this can be easily proved by induction. But I am looking for another approach. How do I prove this without induction? Here this question exists - How does one prove that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}?$. But the only one solution there uses induction. But I am looking for solution other than induction.

• – Nosrati Feb 9 '17 at 13:18
• @MyGlasses Nope... It is not asking to find a prove by induction .... The only answer there uses induction – Rezwan Arefin Feb 9 '17 at 13:27
• Any general statement involving all natural numbers $n$ will at some level rely for proof on "induction" as an axiomatic principle that defines natural numbers. Perhaps you should explain more fully your motive in asking for a proof "without induction". – hardmath Feb 11 '17 at 2:13

## 1 Answer

If we consider $$a_n = \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n}=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}$$ we have: $$a_n^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right) = \frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{2}$$ hence: $$4n a_n^2 \leq \exp\sum_{k=2}^{n}\frac{1}{4k(k-1)} \leq \exp\left(\frac{1}{4}\right) \tag{3}$$ and: $$a_n \leq \sqrt{\frac{1}{4e^{-1/4}n}} \tag{4}$$ is a stronger inequality, since $4e^{-1/4}\approx 3+\frac{1}{9}$.
No induction, just squaring and creative telescoping.

• Wow, very nice! +1 – Zubin Mukerjee Feb 9 '17 at 13:20
• Seems to hard to me :( :'( – Rezwan Arefin Feb 9 '17 at 13:27
• @RezwanArefin V. Hard. This is my attempt, hope it at least has some truth in it. imgur.com/a/KioSR – Cppg Feb 9 '17 at 14:42
• @Cppg Your's are harder than this one :V :V – Rezwan Arefin Feb 9 '17 at 14:46
• This is so elegant ! Thanks for providing answers of this quality. Cheers. – Claude Leibovici Feb 9 '17 at 15:10